AMC 8 · 2017 · #7
Easy mode Grade 4Problem
Picture a 6-digit number where the first three digits and the last three digits are exactly the same, in the same order.
For example, 247247 works: the first three digits are 247, and so are the last three. Another one is 358358.
Call any such 6-digit number Z.
Among the five choices below, one number is guaranteed to be a factor of every such Z. Which one?
Pick an answer.
AMC 8 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A 6-digit number $Z$ has the special form where its first three digits repeat as its last three digits, like $247247$. The problem asks which of the five choices — $11, 19, 101, 111, 1111$ — must always be a factor of $Z$, no matter what those three repeated digits are.
Givens: $Z$ is a 6-digit positive integer; $Z$ has the form $abcabc$ — the first three digits ($abc$) are repeated as the last three digits ($abc$); Example given in the problem: $247247$; Answer choices: (A) $11$, (B) $19$, (C) $101$, (D) $111$, (E) $1111$
Unknowns: Which of the five choices is guaranteed to divide every number of the form $abcabc$
Understand
Restated: A 6-digit number $Z$ has the special form where its first three digits repeat as its last three digits, like $247247$. The problem asks which of the five choices — $11, 19, 101, 111, 1111$ — must always be a factor of $Z$, no matter what those three repeated digits are.
Givens: $Z$ is a 6-digit positive integer; $Z$ has the form $abcabc$ — the first three digits ($abc$) are repeated as the last three digits ($abc$); Example given in the problem: $247247$; Answer choices: (A) $11$, (B) $19$, (C) $101$, (D) $111$, (E) $1111$
Plan
Primary tool: #9 Solve an Easier Related Problem
Secondary: #5 Look for a Pattern, #3 Eliminate Possibilities
Instead of jumping into algebra, we use Tool #9 (Easier Problem) by working with the concrete example $247247$ the problem hands us, plus one or two more $abcabc$ numbers. Tool #5 (Pattern) helps us notice that every $abcabc$ number is exactly $abc \times 1001$ — once we see that pattern, the question reduces to "which choices divide $1001$?". Finally Tool #3 (Eliminate Possibilities) checks the five choices against $1001$ to pick the winner. We deliberately avoid Tool #13 (Algebra) — the place-value pattern is easier to discover by playing with one example than by setting up variables first.
Execute — Answer: A
4.NBT.B.6 Step 1 - Start with the example $247247$ that the problem gives us.
- Divide it by the repeated three-digit block $247$ to see what's left.
💡 Dividing a 6-digit number by a 3-digit number is a Grade 4 long-division skill — no algebra needed.
4.OA.C.5 Step 2 - Test the pattern with another $abcabc$ number to make sure $1001$ wasn't a coincidence.
- Try $315315 \div 315$.
💡 Computing two cases and seeing the same quotient $1001$ is the Grade 4 way of confirming a number pattern before trusting it.
4.NBT.A.2 Step 3 - Explain WHY the quotient is always $1001$, using place value.
- The first three digits $abc$ sit in the hundred-thousands, ten-thousands, and thousands places, which is $1000$ copies of $abc$.
- The last three digits are just $1$ more copy of $abc$.
- So $abcabc = abc \times 1000 + abc = abc \times 1001$.
💡 Reading $abcabc$ by place value — the first $abc$ is worth $1000$ times the second $abc$ — is exactly the Grade 4 multi-digit place-value standard.
4.OA.B.4 Step 4 - Since $Z = \overline{abc} \times 1001$ for every such number, any factor of $1001$ is automatically a factor of $Z$.
- So the real question becomes: which of $11, 19, 101, 111, 1111$ divides $1001$?
- Find the prime factorization of $1001$.
💡 Finding factor pairs of a whole number (here, $1001$) is the Grade 4 factor-pair / prime-or-composite standard.
4.OA.B.4 Step 5 - Check the five choices against $1001$ to eliminate.
- $11$ is one of the prime factors — it divides $1001$.
- The others do not: $1001 \div 19 \approx 52.7$ (not whole), $1001 \div 101 \approx 9.91$, $1001 \div 111 \approx 9.02$, $1001 \div 1111 < 1$.
- Only $11$ survives.
💡 Testing each choice as a divisor of $1001$ is the same Grade 4 factor-pair check applied to a multiple-choice elimination — only the true factor stays standing.
4.NBT.B.6 Start with the example $247247$ that the problem gives us. Divide it by the repe 4.OA.C.5 Test the pattern with another $abcabc$ number to make sure $1001$ wasn't a coinc 4.NBT.A.2 Explain WHY the quotient is always $1001$, using place value. The first three di 4.OA.B.4 Since $Z = \overline{abc} \times 1001$ for every such number, any factor of $100 4.OA.B.4 Check the five choices against $1001$ to eliminate. $11$ is one of the prime fac Review
Reasonableness: Spot-check with two different $abcabc$ numbers. $247247 \div 11 = 22477$ (whole) and $123123 \div 11 = 11193$ (whole) — $11$ works for both. Now try a choice that shouldn't always work, like $19$: $123123 \div 19 \approx 6480.16$ is not whole, so $19$ is not a guaranteed factor. Only $11$ — sitting inside $1001 = 7 \times 11 \times 13$ — survives for every $abcabc$. Answer (A) $\boxed{11}$ holds up.
Alternative: Tool #13 (Convert to Algebra) gives a one-line proof: let $N = \overline{abc}$, then $Z = 1000N + N = 1001N$, so every factor of $1001 = 7 \times 11 \times 13$ divides $Z$. From the choices, only $11$ is among $\{7, 11, 13\}$. The algebra route is shorter, but the pattern-discovery route in our main path makes the $1001$ appear naturally instead of being announced.
CCSS standards used (min grade 4)
4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Dividing the concrete example $247247$ by $247$ to discover that the quotient is $1001$ — uses Grade 4 long division (the extra digits beyond four-digit dividends are handled by the same algorithm).)4.OA.C.5Generate a number or shape pattern following a given rule (Confirming with a second example ($315315 \div 315 = 1001$) that every $abcabc$ number follows the same pattern before trusting the rule.)4.NBT.A.2Read and write multi-digit whole numbers and compare using symbols (Reading the 6-digit number $abcabc$ by place value, recognizing the first $abc$ block is worth $1000$ times the second $abc$ block, which gives $abcabc = abc \times 1001$.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Factoring $1001 = 7 \times 11 \times 13$ and then testing each multiple-choice option ($11, 19, 101, 111, 1111$) as a possible divisor of $1001$.)
⭐ This AMC 8 problem only needs Grade 4 place value and factor pairs you already know — once you see that $abcabc = abc \times 1001$, it's just asking which choice divides $1001$!
⭐ This AMC 8 problem only needs Grade 4 place value and factor pairs you already know — once you see that $abcabc = abc \times 1001$, it's just asking which choice divides $1001$!
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